I see two questions here. The first is why self-inductance is not considered when solving Faraday's law problems, and the second is why an EMF can ever produce a current in a circuit with non-zero self-inductance. I will answer both of these in turn.
1. Why self-inductance is not considered when solving Faraday's law problems
Self inductance should be considered, but is left out for simplicity. So for example, if you have a planar circuit with inductance $L$, resistance $R$, area $A$, and there is a magnetic field of strength $B$ normal to the plane of the circuit, then the EMF is given by $\mathcal{E}=-L \dot{I} - A \dot{B}$.
This means, for example, that if $\dot{B}$ is constant, then, setting $IR=\mathcal{E}$, we find $\dot{I} = -\frac{R}{L} I - \frac{A}{L} \dot{B}$. If the current is $0$ at $t=0$, then for $t>0$ the current is given by $I(t)=-\frac{A}{R} \dot{B} \left(1-\exp(\frac{-t}{L/R}) \right)$. At very late times $t \gg \frac{L}{R}$, the current is $-\frac{A \dot{B}}{R}$, as you would find by ignoring the inductance. However, at early times, the inductance prevents a sudden jump of the current to this value, so there is a factor of $1-\exp(\frac{-t}{L/R})$, which causes a smooth increase in the current.
2. Why an EMF can ever produce a current in a circuit with non-zero self-inductance.
You are worried that EMF caused by the circuit's inductance will prevent any current from flowing. Consider the planar circuit as in part one, and suppose there is a external emf $V$ applied to the circuit (and no longer any external magnetic field). The easiest way to see that current will flow is by making an analogy with classical mechanics: the current $I$ is analogous to a velocty $v$; the resistance is analogous to a drag term, since it represents dissipation; the inductance is like mass, since the inductance opposes a change in the current the same way a mass opposes a change in velocity; and the EMF $V$ is analogous to a force. Now you have no problem believing that if you push on an object in a viscous fluid it will start moving, so you should have no problem believing that a current will start to flow.
To analyze the math, all we have to do is replace $-A \dot{B}$ by $V$ in our previous equations, we find the current is $I(t) = \frac{V}{R} \left(1-\exp(\frac{-t}{L/R}) \right)$, so as before the current increases smoothly from $0$ to its value $\frac{V}{R}$ at $t=\infty$.
The inductance of a coaxial cable does depend on the number of coils. But in addition to that, a coaxial cable will have a nonzero inductance even if it is perfectly straight, because there is a magnetic field inside the cable whenever it is carrying current, and this takes energy to set up, so you cannot instantaneously increase the current from zero to $I$ without performing work to set up the magnetic field.
Your book is calculating this straight-wire self inductance.If you coil up the cable then you will have that as well, but it will be very small because the cable has zero net current, and the magnetic field inside the loop will be very small - due to small asymmetries and imperfections.
In practice, you never really coil up coaxial wire, which is used for communications because of it shielding properties. The only reason you intentionally coil up wires and run current through them is when you actively want the inductance to be large, and coaxial cable will be terrible at that.
Best Answer
I agree that this seems mysterious on first meeting.
Multiplication by $n$ is shown to be correct by something called Stokes's Theorem, which you won't have met yet, and which lets us translate the basic equations of electromagnetism (called their Maxwell Equations) between their local and "spread out" forms.
But at an easier level, think of a very long, time varying flux tube, like the one I've drawn below.
Now imagine separate loops around the flux tube. Each of these loops has flux $\phi$ through them, and each separately will have an EMF $-\frac{{\rm d}}{{\rm d}\,t}\phi$ between their ends.
You link them together in series, and it is just like linking voltage sources, or batteries together in series. The total EMF across the series cells is $-n\,\frac{{\rm d}}{{\rm d}\,t}\phi$. Now imagine bringing the loops together in a neatly wound solenoid. Electrically nothing changes: it doesn't matter if you put long, zero resistance wires between batteries in series, their EMFs still add in exactly the same way.
Since the whole purpose of an inductor is react to a change in current with a back EMF, the physics doesn't change if you state the equation $V = -n\,\frac{{\rm d}}{{\rm d}\,t}\phi$ as $L I = -n\,\phi$; you simply differentiate this one to get the form I just derived.